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4-2.Friction
hard
The rear side of a truck is open and a box of $40 \;kg$ mass is placed $5 \,m$ away from the open end as shown in Figure. The coeffictent of friction between the box and the surface below it is $0.15 .$ On a stratght road, the truck starts from rest and accelerates with $2\; m s ^{-2} .$ At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).
Option A
Option B
Option C
Option D
Solution
Mass of the box, $m=40 \,kg$
Coefficient of friction, $\mu=0.15$
Initial velocity, $u=0$
Acceleration, $a=2 \,m / s ^{2}$
Distance of the box from the end of the truck, $s^{\prime}=5\, m$
As per Newton's second law of motion, the force on the box caused by the accelerated motion of the truck is given by:
$F=m a$
$=40 \times 2=80 \,N$
As per Newton's third law of motion, a reaction force of $80 \,N$ is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction $f,$ acting between the box and the floor of the truck. This force is given by:
$f=\mu m g$
$=0.15 \times 40 \times 10=60\, N$
$\therefore$ Net force acting on the block:
$F_{\text {net }}=80-60=20\, N$ backward
The backward acceleration produced in the box is given by:
aback $\quad=\frac{F_{\text {mat }}}{m}=\frac{20}{40}=0.5 \,m / s ^{2}$
Using the second equation of motion, time $t$ can be calculated as:
$s^{\prime}=u t+\frac{1}{2} a_{ back } t^{2}$
$5=0+\frac{1}{2} \times 0.5 \times t^{2}$
$\therefore t=\sqrt{20} \,s$
Hence, the box will fall from the truck after $\sqrt{20}$ $s$ from start.
The distance $s$, travelled by the truck in $\sqrt{20} \,s$ is given by the relation:
$s=u t+\frac{1}{2} a t^{2}$
$=0+\frac{1}{2} \times 2 \times(\sqrt{20})^{2}$
$=20\, m$
Coefficient of friction, $\mu=0.15$
Initial velocity, $u=0$
Acceleration, $a=2 \,m / s ^{2}$
Distance of the box from the end of the truck, $s^{\prime}=5\, m$
As per Newton's second law of motion, the force on the box caused by the accelerated motion of the truck is given by:
$F=m a$
$=40 \times 2=80 \,N$
As per Newton's third law of motion, a reaction force of $80 \,N$ is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction $f,$ acting between the box and the floor of the truck. This force is given by:
$f=\mu m g$
$=0.15 \times 40 \times 10=60\, N$
$\therefore$ Net force acting on the block:
$F_{\text {net }}=80-60=20\, N$ backward
The backward acceleration produced in the box is given by:
aback $\quad=\frac{F_{\text {mat }}}{m}=\frac{20}{40}=0.5 \,m / s ^{2}$
Using the second equation of motion, time $t$ can be calculated as:
$s^{\prime}=u t+\frac{1}{2} a_{ back } t^{2}$
$5=0+\frac{1}{2} \times 0.5 \times t^{2}$
$\therefore t=\sqrt{20} \,s$
Hence, the box will fall from the truck after $\sqrt{20}$ $s$ from start.
The distance $s$, travelled by the truck in $\sqrt{20} \,s$ is given by the relation:
$s=u t+\frac{1}{2} a t^{2}$
$=0+\frac{1}{2} \times 2 \times(\sqrt{20})^{2}$
$=20\, m$
Standard 11
Physics
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